Answer
The inequality is valid for values less than 3 and values more than 7 (including them) i.e. $(-\infty,3]\cap [7,\infty)$.
Work Step by Step
First, we are going to move everything to the left side and simplify:
$\dfrac{x+1}{x-3}\leq2$
$\dfrac{x+1}{x-3}-2\leq0$
$\dfrac{x+1}{x-3}-\dfrac{2(x-3)}{x-3}\leq0$
$\dfrac{x+1-(2x-6)}{x-3}\leq0$
$\dfrac{-x+7}{x-3}\leq0$
Now, we find critical points by equating the numerator and denominator to zero:
$x-3=0$
$-x+7=0$
There are two critical points:
$x_1-3=0\rightarrow x_1=3$
$-x_2+7=0\rightarrow x_2=7$
Next, we are going to take three values: one less than 3; one between 3 and 7; and one more than 7 to test in the original equation and check if the inequality is true or not:
First test with a value less than 3:
$\dfrac{2+1}{2-3}\leq2$
$\dfrac{3}{-1}\leq2$
$-3\leq2 \rightarrow \text{ TRUE}$
Second test with a value between 3 and 7:
$\dfrac{4+1}{4-3}\leq2$
$\dfrac{5}{1}\leq2$
$5\leq2 \rightarrow \text{ FALSE}$
Third test with a value more than 7:
$\dfrac{13+1}{13-3}\leq2$
$\dfrac{14}{10}\leq2$
$1.4\leq2 \rightarrow \text{ TRUE}$
These tests show that the inequality $\dfrac{x+1}{x-3}\leq2$ is valid for values less than 3 and values more than 7 (including them) i.e. $(-\infty,3]\cap [7,\infty)$